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Question

If a2+b2+c2=1, then bc+ca+ab lies in the interval


A

[12,3]

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B

[-1,2]

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C

[12,1]

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D

[1,12]

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Solution

The correct option is C

[12,1]


We have a2+b2+c2+2(bc+ca+ab)=(a+b+c)20
1+2(bc+ca+ab)0bc+ca+ab12

Also since A.M. G.M., for a, b, cwe get
b2+c22b2c2=bc,c2+a22c2a2=ca and
a2+b22a2b2=ab

Add all three inequalities to get
a2+b2+c2ab+bc+ca
1ab+bc+ca

So, the interval is [12,1].


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